TITLE: Why not define $|v| = -1$?
QUESTION [5 upvotes]: I was wondering why if we have $i^2 = -1$, why not have a "number" $v$ such that $|v| = -1$?  Does anything interesting arise from considering this system?
The only thing I could come up with was:
$$ |a+bv| \le |a| + |bv| = a - b $$

REPLY [4 votes]: First off, it wouldn't be very interesting to have a norm that was negative for all nonzero vectors in a vector space -- that would just be the same old norm we already know, with a minus in front of it.
So suppose we have vectors $v, w$ with $|v|<0$ and $|w|>0$. Let's further assume that $v$ and $w$ are linearly independent. If they are not, assuming that our vector space is at least two dimensional, let $q$ be some vector that is linearly independent of $v$ (and this also $w$). If $|q|>0$ then use $q$ as the new $w$; otherwise use $q$ as the new $v$.
Now consider the function
$$ f(t) = |(1-t)v + tw| $$
This is certainly something that looks like it ought to be continuous on $[0,1]$, but $f(0)=|v|<0$ and $f(1)=|w|>0$, so by the intermediate value theorem, there should be some $t$ such that $|(1-t)v+tw|=0$. And because $v$ and $w$ are linearly independent, $(1-t)v+tw\ne 0$.
Now we have a nozero vector with zero norm, and that wreaks all kinds of havoc for most of the things we want to use norms for.

Bonus chatter: There are in fact spaces with norm-like things $\|{\cdot}\|$ such that $\|v\|^2$ is always real, but can be either positive or negative(!). A prominent example is Minkowski space, which forms the "spacetime" of Special Relativity. Here we do indeed have nonzero vectors with zero $\|v\|$, which play a special role in the theory -- but for that very reason they don't satisfy many of the properties the usual length of a vector has.